Solve the inequality $\sqrt x+\sqrt{x+1}>\sqrt 3$.

Question

Solve the inequality $\sqrt x+\sqrt{x+1}>\sqrt 3$.

I want to make sure my method is correct:

The condition is that $x\geq 0$

$x+2\sqrt x\times \sqrt{x+1} +x+1>3$

$2\sqrt{x(x+1)}>2-2x$

$4x(x+1)>4-8x+4x^{2}$

$4x^{2}+4x>4-8x+4x^{2}$

$12x>4$

$x>\frac{1}{3}$

$x\in (\frac{1}{3},\infty)$

I know my final solution is fine, but is everything written properly? Should I put $\iff$ at the beginning of each row?

Answer 1

Your answer is fine and correct, and you can write it more concise with key steps. In the end, you can simply write $x > 1/3$ without re-write it as $x \in (1/3, \infty)$.

Answer 2

You've made a mistake by assuming that $2 - 2x \ge 0$, when squaring in the second row. To fix this consider the two cases when $x < 1$ and $x \ge 1$.

This will enable you to add the $\iff$ signs, which you're required, as otherwise you have proven that $\sqrt{x} + \sqrt{x+1} > \sqrt{3} \implies x > \frac 13$ instead of $x > \frac 13 \implies \sqrt{x} + \sqrt{x+1} > \sqrt{3}$

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Assume that $x > 1$. Then we have that $2-2x < 0 < 2\sqrt{x(x+1)}$, so the inequality is true for any $x \ge 1$.

On the other side when $x < 1$ you can continue in your way and you will get that $x \in \left(\frac 13,1\right)$. Now combining the answers you will get that the solution set is $\left(\frac 13,\infty\right)$

Answer 3

We have

$x\geq 0$ .

then

$\sqrt{x}+\sqrt{x+1}>\sqrt{3} \implies$

$\sqrt{x+1}>\sqrt{3}-\sqrt{x} \implies $

$x+1>3+x-2\sqrt{3x} \implies$

$\sqrt{3x}>1 \implies x>\frac{1}{3}$

In the other direction,

$x>\frac{1}{3} \implies$

$\sqrt{x+1}+\sqrt{x}>\sqrt{\frac{4}{3}}+\sqrt{\frac{1}{3}} =\sqrt{3}$

Qed.

Answer 4

Over its domain (the set of non-negative real numbers) the function $f(x)=\sqrt{x}+\sqrt{x+1}$ is an increasing function, since it is the sum of two non-negative increasing functions. It follows that $f(x)>\sqrt{3}$ holds as soon as $x>x_0$, where $x_0$ is the only positive number such that $$ \sqrt{x_0}+\sqrt{x_0+1} = \sqrt{3}.\tag{1} $$ $x_0=\frac{1}{3}$ is clearly a solution of $(1)$, hence the given inequality holds for $\color{red}{\large x>\frac{1}{3}}$.

Answer 5

Note that $\sqrt{\frac13}+\sqrt{\frac13+1}=\sqrt{3}$ . The inequality follows trivially.